In a series RL AC circuit, the inductive reactance and resistance are equal in magnitude (XL = 100 Ω and R = 100 Ω). What is the magnitude of the circuit impedance Z?

Introduction:
Determining the impedance of a series RL circuit is a fundamental AC analysis skill. Because resistance and reactance are orthogonal (one dissipative, the other reactive), they combine vectorially rather than arithmetically. This question checks your ability to compute the magnitude of impedance when the resistor and inductor have equal magnitudes (R = XL).

Given Data / Assumptions:

• Series RL circuit (single loop).
• R = 100 Ω.
• XL = 100 Ω.
• Steady-state sinusoidal excitation; use RMS quantities.

Concept / Approach:
For a series RL circuit, impedance magnitude is found using the Pythagorean relationship because R and XL are at 90 degrees in the phasor plane: Z = sqrt(R^2 + XL^2). When R = XL, Z becomes R * sqrt(2).

Step-by-Step Solution:
1) Write the impedance formula for series RL: Z = sqrt(R^2 + XL^2).2) Substitute R = 100 Ω and XL = 100 Ω.3) Compute: Z = sqrt(100^2 + 100^2) = sqrt(10,000 + 10,000) = sqrt(20,000).4) Evaluate: sqrt(20,000) ≈ 141.4 Ω.

Verification / Alternative check:
Use the equality case: Z = R * sqrt(2) = 100 * 1.414 ≈ 141.4 Ω, confirming the result quickly without full arithmetic.

Why Other Options Are Wrong:
14.14 Ω: misses a factor of 10; likely a decimal-place error.100 Ω: wrongly adds vector components linearly along one axis.200 Ω: simple sum R + XL; not valid for orthogonal phasors.

Common Pitfalls:
Adding R and XL directly, or confusing series and parallel combinations. Always remember that resistive and inductive components in series combine as a right triangle in the phasor diagram.

Final Answer:
141.4 Ω